Ma connection at home is again functioning. I am happy to have solved the problem rather quickly.
On 22 Jul 2009, at 13:54, David Nyman wrote: > > 2009/7/22 Bruno Marchal <[email protected]>: > >> Ma connection at home is no functioning. > > As a linguistic aside, Bruno has cleverly expressed the above > statement in perfect Glaswegian (i.e. the spoken tongue of Glasgow, > Scotland - my home town). Indeed I am following an intense summer school in Glaswegian. You thought you could make fun of the poor disconnected one? > Other well-known examples are: "Is'arra > marra on yer barra Clarra?" (Is that large vegetable on your barrow a > marrow, Clara?); and "Gie's'a sook on yer soor ploom" (Let me taste > the "sour plum" (a globular sweet-sour confection) that you are > presently sucking). > > Perhaps he intends to continue further in this vein? Once, on a list, someone thought I was using slang from New-York! Now Glaswegian! I am afraid I am just writing to quickly, and then when I read myself I concentrate so much on the meaning ... Most of the time I see the spelling errors when I read my mail, never when I send it. Sorry sorry sorry ... Take care of the sense and the spelling will take care of itself. Well *that* does not work! Bruno > > > David ;-) > >> Hi, >> >> Ma connection at home is no functioning. So I am temporarily >> disconnected. I hope I will be able to solve that problem. I am >> sending >> here some little comments from my office. >> >> I include some more material for Kim and Marty, and others, just to >> think about, in case I remain disconnected for some time. Sorry. >> >> Bruno >> >> Le 22-juil.-09, à 10:27, Torgny Tholerus a écrit : >> >>> >>> Rex Allen skrev: >>>> Brent, >>>> >>>> So my first draft addressed many of the points you made, but it >>>> that >>>> email got too big and sprawling I thought. >>>> >>>> So I've focused on what seems to me like the key passage from your >>>> post. If you think there was some other point that I should have >>>> addressed, let me know. >>>> >>>> So, key passage: >>>> >>>> >>>>> Do these mathematical objects "really" exist? I'd say they have >>>>> logico-mathematical existence, not the same existence as tables >>>>> and >>>>> chairs, or quarks and electrons. >>>>> >>>> >>>> So which kind of existence do you believe is more fundamental? >>>> Which >>>> is primary? Logico-mathematical existence, or quark existence? Or >>>> are they separate but equal kinds of existence? >>>> >>>> >>> >>> The most general form of existence is: All mathematical possible >>> universes exist. Our universe is one of those mathematical possible >>> existing universes. >> >> This is non sense. Proof: see UDA. Or interrupt me when you have an >> objection in the current explanation. I have explained this many >> times, >> but the notion of universe or mathematical universe just makes no >> sense. The notion of "our universe" is too far ambiguous for just >> making even non sense. >> >> I could say the same to Brent. First I don't think it makes sense to >> say that epistemology comes before ontology, given that the ontology, >> by definition, in concerned with what we agree exist independently of >> the observer/knower ... Then what you say contradict the results in >> the >> computationalist theory, where the appearances of universe emerges >> from >> the collection of all computations >> >> BTW, thanks to Brent for helping Marty. >> >> Rex, when you say: >> >>> I would say that most people PERCEIVE logico-mathematical objects >>> differently than they perceive tables and chairs, or quarks and >>> electrons. But this doesn't tell us anything about whether these >>> things really have different kinds of existence. That we perceive >>> them differently is just an accident of fate. >> >> We perceive them differently because "observation" is a different >> modality of self-reference than "proving". It has nothing to do with >> accident or fate. The comp physics is defined by what is invariant, >> from the "observation" point of view of universal machine. Later this >> will shown to be given by the 3th, 4th, and 5th hypostases. >> >> ==== math lesson ==== (2 posts): >> >> Hi, >> >> I wrote: >> << >> The cardinal of { } = 0. All singletons have cardinal one. All pairs, >> or doubletons, have cardinal two. >> >> Problem 1 has been solved. They have the same cardinal, or if you >> prefer, they have the same number of elements. The set of all subsets >> of a set with n elements has the same number of elements than the set >> of all strings of length n. >> >> Let us write B_n for the sets of binary strings of length n. So, >> >> B_0 = { } >> B_1 = {0, 1} >> B_2 = {00, 01, 10, 11} >> B_3 = {000, 010, 100, 110, 001, 011, 101, 111} >> >> We have seen, without counting, that the cardinal of the powerset >> of a >> set with cardinal n is the same as the cardinal of B_n. >> >> >> >> >> And now the killing question by the sadistic math teacher: >> >> What is the cardinal, that is, the number of element, of B_0, that is >> the set of strings of length 0. >> >> The student: let see, you wrote above B_0 = { },, and you were kind >> enough to recall that the cardinal of { } is zero (of course, there >> is >> zero element in the empty set). So the cardinal of B_0 is zero. >> 'zero" >> said the student. >> >> 'zero' indeed, said the teacher, but it is your note. You are wrong. >> >> B_0 is not empty! It *looks* empty, but beware the appearance, it >> looks >> empty because it contains the empty string, which, if you remember >> some >> preceding post is invisible (even under the microscope, telescope, >> radioscope, ..). >> >> A solution could have been to notate the empty string by a symbol >> like >> "_", and write all sequences "0111000100" starting from "_": >> _0111000100, with rules __ = _, etc. Then B_0 = { _ }, B_1 = {_0, >> _1}, etc. But this is too much notation. >> >> >> And now the time has come for contrition when the teacher feels >> guilty! >> >> Ah..., I should have written directly something like >> >> B_0 = { _ }, with _ representing the empty sequence. >> B_1 = {0, 1} >> B_2 = {00, 01, 10, 11} >> B_3 = {000, 010, 100, 110, 001, 011, 101, 111} >> >> OK? >> >> Remember we have seen that the cardinal of the powerset of a set >> with n >> elements is equal to the cardinal of B_n, is equal to 2^n. >> >> The cardinal of B_0 has to be equal to to 2^0, which is equal to one. >> Why? >> >> if a is a number, usually, a^n is the result of effectuating (a >> times a >> times a time a ... times a), with n occurences of a. For example: >> 2^3 = >> 2x2x2 = 8. >> >> so a^n times a^m is equal to a^(n+m) >> >> This extends to the rational by defining a^(-n) by 1/a^n. In that >> case >> a^(m-n) = a^m/a^n. In particular a^m/a^m = 1 (x/x = 1 always), and >> a^m/a^m = a^(m-m) = a^0. So a^0 = 1. So in particular 2^0 = 1. >> >> But we will see soon a deeper reason to be encouraged to guess that >> a^0 >> = 1, but for this we need to define the product and the >> exponentiation >> of sets. if A is a set, and B is a set: the exponential B^A is a very >> important object, it is where the functions live. >> >> Take it easy, and ask. Verify the statements a^n/a^m = a^(n-m), >> with n >> = 3 and m = 5. What is a*a*a/a*a*a*a*a "/" = division, and * = >> times). >> >> Bruno >> >> >> >> http://iridia.ulb.ac.be/~marchal/ >> >> ----------------- >> >> Hi, >> >> I am thinking aloud, for the sequel. >> >> >> There will be a need for a geometrical and number theoretical >> interlude. >> >> Do you know what is a periodic decimal? >> >> Do you know that a is periodic decimal if and only if it exists n and >> m, integers, such that a = n/m. And that for all n m, n/m is a >> periodic decimal? >> >> Could you find n and m, such that >> 12.95213213213213213213213213213213213213 ... (= 12.95 213 213 ...) >> >> Solution: >> >> Let k be a name for 12.95213213213213213213213213213213213213213 ... >> >> Let us multiply k by 100 000. >> >> 100 000k = 1295213.213213213213213213213213213213213213 ... = >> 1295213 >> + 0.213213213 ... >> >> Let us multiply k by 100 >> >> 100k = 1295.213213213213... = 1295 + 0.213213213213213.. >> >> >> We have 100000k - 100k = 1295213 + 0.213213213... - 1295 >> - 0.213213213... = 1295213 - 1295 = 1293918 >> >> So 99900k = 1293918 >> >> Dividing by 99900 the two sides of the egality we get: >> >> k = 1293918/99900 >> >> We have n and m such that k = n/m = 12.95213213213213213... >> n = 1293918, and m = 99900. >> >> This should convince you that all periodic decimal are fractions. >> >> Exercice: find two numbers n and m such that n/m = >> 31,2454545454545454545... = 31, 2 45 45 45 45 ... >> >> >> Convince yourself that for all n and m, n/m gives always a periodic >> decimal.(hint: when n is divided by m, m bounds the number of >> possible >> remainders). >> >> And now geometry (without picture, do them). >> >> Do you know that the length of the circle divided by its diameter is >> PI? (PI = 3.141592...) >> Do you know that the length of the square divided by its diagonal is >> the square root of 2? (sqrt(2)= 1,414213562...) >> - can you show this? >> - can you show this without Pythagorus theorem? (like in Plato!) >> >> Do you know if it exists n and m such that n/m = the square root of 2 >> (relation with incommensurability) >> Do you know if the Diophantine equation x^2 = 2y^2 has a solution? >> >> No. >> I think I will prove this someday, if only to have an example of >> simple, yet non trivial, proof. >> >> This entails that the sqaure root of 2 cannot be equal to any >> fraction >> n/m. >> And it means the square root of 2 is a non periodic decimal. (its >> decimal will provide a good example of a non trivial computable >> function). >> >> Bruno >> >> http://iridia.ulb.ac.be/~marchal/ >> >>> >> > > > http://iridia.ulb.ac.be/~marchal/ --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to the Google Groups "Everything List" group. To post to this group, send email to [email protected] To unsubscribe from this group, send email to [email protected] For more options, visit this group at http://groups.google.com/group/everything-list?hl=en -~----------~----~----~----~------~----~------~--~---
